Ad ơi báo TT có đăng lời giải rồi, nhưng trong lời giải họ sửa từ \(\sqrt{x^2+7x-1}\) thành \(\sqrt{x^2+7x-4}\).

\(...\Leftrightarrow2019\left(x-1\right)^2+2020\left(y+1\right)^2=0\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)

Ta có \(\sqrt{1+a^2}+\sqrt{2a}\le\sqrt{2\left(1+a^2+2a\right)}=\sqrt{2}\left(a+1\right)\).

Tương tự \(\sqrt{1+b^2}+\sqrt{2b}\le\sqrt{2}\left(b+1\right)\); \(\sqrt{1+c^2}+\sqrt{2c}\le\sqrt{2}\left(c+1\right)\).

Lại có \(\left(2-\sqrt{2}\right)\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\le\left(2-\sqrt{2}\right)\sqrt{3\left(a+b+c\right)}\le3\left(2-\sqrt{2}\right)\).

Do đó \(B\le\sqrt{2}\left(a+b+c+3\right)+3\left(2-\sqrt{2}\right)\le6\sqrt{2}+6-3\sqrt{2}=3\sqrt{2}+6\).

Dấu "=" xảy ra khi a = b = c = 1.

Ta có \(F=sin^2\dfrac{\pi}{6}+...+sin^2\pi=\left(sin^2\dfrac{\pi}{6}+sin^2\dfrac{5\pi}{6}\right)+\left(sin^2\dfrac{2\pi}{6}+sin^2\dfrac{4\pi}{6}\right)+\left(sin^2\dfrac{3\pi}{6}+sin^2\pi\right)=\left(sin^2\dfrac{\pi}{6}+cos^2\dfrac{\pi}{6}\right)+\left(sin^2\dfrac{2\pi}{6}+cos^2\dfrac{2\pi}{6}\right)+\left(1+0\right)=1+1+1=3\)

Ta có \(\overline{abcd}⋮4\Leftrightarrow1000a+100b+10c+d⋮4\Leftrightarrow4\left(250a+25b+2c\right)+2c+d⋮4\Leftrightarrow2c+d⋮4\).